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whitney_golden Jan 19, 2026 • 0 views

Pre-Calculus Bearing Problems: Finding Displacement and Total Distance

Hey everyone! 👋 I'm stuck on these pre-calculus bearing problems. They always seem to involve displacement and total distance. Can someone explain how to solve them in a way that actually makes sense? 🙏 It feels like I'm just memorizing steps without understanding what I'm doing. Any help is appreciated!
🧮 Mathematics

1 Answers

✅ Best Answer

📚 Understanding Bearing Problems in Pre-Calculus

Bearing problems in pre-calculus involve finding distances and directions using angles. These problems often involve concepts of trigonometry, especially the sine and cosine functions, and the Law of Sines and Law of Cosines. They are commonly used in navigation, surveying, and aviation to determine positions and distances.

🧭 History and Background

The concept of bearings has ancient roots, primarily used in navigation. Early navigators used celestial bodies and rudimentary instruments to determine their direction. As technology advanced, more accurate instruments like the magnetic compass and later GPS systems were developed. The mathematical foundations for solving bearing problems developed alongside trigonometry, with significant contributions from Greek and Islamic scholars.

📐 Key Principles

  • 🧭 Bearing Definition: Bearing is the angle, measured clockwise, from true north to a line of travel. It is typically expressed in degrees.
  • 📐 Angle of Elevation and Depression: These angles are formed between a horizontal line and the line of sight to an object above (elevation) or below (depression).
  • 📏 Displacement: The straight-line distance and direction from the starting point to the ending point. It's a vector quantity.
  • 🚶 Total Distance: The total length of the path traveled, regardless of direction. It's a scalar quantity.
  • 💡 Trigonometric Functions: Sine, cosine, and tangent are used to relate angles and sides of right triangles.
  • 📜 Law of Sines: In any triangle, the ratio of the length of a side to the sine of its opposite angle is constant: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$.
  • 📝 Law of Cosines: Relates the sides and angles in any triangle: $c^2 = a^2 + b^2 - 2ab \cos C$.

🌍 Real-World Examples

Let's consider a few examples:

Example 1: Finding Displacement

A ship sails 50 km on a bearing of 040° and then sails 70 km on a bearing of 110°. Find the displacement.

Solution:

1. Draw a Diagram: Draw the bearings and distances on a coordinate plane.

2. Resolve into Components:

  • For the first leg:
  • Eastward component: $50 \sin(40°) \approx 32.14$ km
  • Northward component: $50 \cos(40°) \approx 38.30$ km
  • For the second leg:
  • Eastward component: $70 \sin(110°-90°) = 70 \sin(20°) \approx 23.94$ km
  • Southward component: $70 \cos(20°) \approx 65.78$ km

3. Calculate Resultant Components:

  • Total Eastward: $32.14 + 23.94 \approx 56.08$ km
  • Total Northward: $38.30 - 65.78 \approx -27.48$ km (Southward)

4. Find Magnitude and Direction:

  • Magnitude (Displacement): $\sqrt{(56.08)^2 + (-27.48)^2} \approx 62.2$ km
  • Direction (Bearing): $\arctan(\frac{56.08}{-27.48}) \approx -63.9°$. Since it's in the fourth quadrant, the bearing is $360 - 63.9 \approx 296.1°$

Therefore, the displacement is approximately 62.2 km on a bearing of 296.1°.

Example 2: Finding Total Distance

A hiker walks 3 km east, then 4 km north, and then 5 km west. Find the total distance traveled and the displacement.

Solution:

  • Total Distance: $3 + 4 + 5 = 12$ km
  • Displacement: The hiker walked 2 km west and 4 km north from the starting point. Displacement magnitude = $\sqrt{2^2 + 4^2} = \sqrt{20} \approx 4.47$ km. The direction can be calculated using arctangent to find the angle.

Example 3: Using Law of Cosines

Two ships leave a port at the same time. Ship A travels at 20 km/h on a bearing of 030°, and Ship B travels at 25 km/h on a bearing of 120°. After 2 hours, how far apart are the ships?

Solution:

1. Calculate Distances:

  • Distance A: $20 \text{ km/h} \times 2 \text{ h} = 40$ km
  • Distance B: $25 \text{ km/h} \times 2 \text{ h} = 50$ km

2. Find the Angle Between the Paths:

  • Angle: $120° - 30° = 90°$

3. Apply Law of Cosines:

  • $c^2 = a^2 + b^2 - 2ab \cos C$
  • $c^2 = 40^2 + 50^2 - 2(40)(50) \cos 90°$
  • $c^2 = 1600 + 2500 - 0$
  • $c^2 = 4100$
  • $c = \sqrt{4100} \approx 64.03$ km

The ships are approximately 64.03 km apart after 2 hours.

✍️ Practice Quiz

Test your understanding with these practice problems:

Question 1:

A plane flies 150 km on a bearing of 070°, then 200 km on a bearing of 160°. Find the displacement.

Question 2:

A boat travels 8 km east, then 6 km south. Find the total distance and displacement.

Question 3:

Two cars leave an intersection at the same time. Car A travels at 60 km/h north, and Car B travels at 80 km/h east. How far apart are they after 1.5 hours?

Question 4:

A hiker walks 5 km on a bearing of 035° and then 3 km on a bearing of 125°. Find the displacement.

Question 5:

A ship sails 100 km on a bearing of 210°, then 50 km on a bearing of 300°. Calculate the displacement.

Question 6:

A person walks 4 km west, then 7 km north, and then 2 km east. Determine the total distance and displacement.

Question 7:

Two airplanes depart from the same airport. Plane A flies at 400 km/h on a bearing of 050°, and Plane B flies at 500 km/h on a bearing of 330°. How far apart are they after 3 hours?

✅ Conclusion

Mastering bearing problems involves a good understanding of trigonometry, vectors, and coordinate geometry. By drawing diagrams and breaking down movements into components, you can effectively solve these problems in various real-world scenarios. Keep practicing, and you'll become proficient in no time!

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